The Longest Straight（FZUoj2216)

 Problem 2216 The Longest Straight

Accept: 82    Submit: 203
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

ZB is playing a card game where the goal is to make straights. Each card in the deck has a number between 1 and M(including 1 and M). A straight is a sequence of cards with consecutive values. Values do not wrap around, so 1 does not come after M. In addition to regular cards, the deck also contains jokers. Each joker can be used as any valid number (between 1 and M, including 1 and M).

You will be given N integers card[1] .. card[n] referring to the cards in your hand. Jokers are represented by zeros, and other cards are represented by their values. ZB wants to know the number of cards in the longest straight that can be formed using one or more cards from his hand.

 Input

The first line contains an integer T, meaning the number of the cases.

For each test case:

The first line there are two integers N and M in the first line (1 <= N, M <= 100000), and the second line contains N integers card[i] (0 <= card[i] <= M).

 Output

For each test case, output a single integer in a line -- the longest straight ZB can get.

 Sample Input

2

7 11

0 6 5 3 0 10 11

8 1000

100 100 100 101 100 99 97 103

 Sample Output

5

3

 Source

第六届福建省大学生程序设计竞赛-重现赛（感谢承办方华侨大学）

思路：尺取法，先将哪些点是否出现过标记，并记录0的个数，然后再前缀数组统计1到M中的未出现的数字，

在用尺取法跑一遍区间取最长就可以了。

#include<stdio.h>#include<algorithm>#include<iostream>#include<stdlib.h>#include<string.h>using namespace std;int aa[100005];int bb[100005];int main(void){    int n,i,j,k,p,q;    scanf("%d",&k);    while(k--)    {        memset(aa,0,sizeof(aa));        memset(bb,0,sizeof(bb));        scanf("%d %d",&p,&q);int ma=0;        for(i=0; i<p; i++)        {            scanf("%d",&n);            aa[n]++;            if(ma<n)            {                ma=n;            }        }        int cnt=aa[0];        bb[0]=0;ma=min(ma+cnt,q);        for(i=1; i<=q; i++)        {            if(!aa[i])            {                bb[i]=bb[i-1]+1;            }            else            {                bb[i]=bb[i-1];            }        }        int ll=1;        int rr=1;        int maxx=0;        while(rr<=ma&&ll<=rr+1)        {            while(bb[rr]-bb[ll-1]<=cnt&&rr<=ma)            {                rr++;            }            maxx=max(maxx,(rr-ll));            ll++;        }        printf("%d\n",maxx);    }    return 0;}

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